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Estimation
by Gail Englert

A response to the question:

What is the difference between methods for estimation?

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I am currently working on a project to improve math test scores on the Stanford 9 Achievement Test for grades 3, 4, 5.

The Compendium of Instructional Objectives lists these objective categories for estimation:

  1. Reasonableness
  2. Front-End Estimation
  3. Estimation with Whole Numbers: Clustering
  4. Estimation with Whole Numbers: Compatible Numbers

I need good "Child-friendly" and/or "Teacher-friendly" explanations regarding the differences.

Any assistance will be appreciated, as I will be inservicing other teachers and giving direct student instruction.

Steve

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Dear Steve,

1. Reasonableness

This is just having an awareness of whether an answer could be correct or not. For example, if you are multiplying two decimals, 0.5 and 12.6, and are given the product 63, red flags should go up because it is not reasonable to think that "half of 12.6" could be 63.

2. Front-End Estimation

This is taking the front parts of the amounts and computing only with them, then adjusting your answer using a quick estimate of the rest of the numbers. For example, if you were to add 545 + 391 + 143 you would have 5 hundreds plus 3 hundreds plus 1 hundred, which is 9 hundreds. Then you look at the rest, and see that 45 and 43 is about a hundred more, and so is 91, so your new sum is 9 hundreds plus two more hundreds, or 1100.

3. Estimation with Whole Numbers: Clustering

I believe this is a way of rearranging numbers as you add to get easier numbers to work with. So, if you were adding 17 + 38 + 43 you might want to add the 17 and 43 first. I am not sure about this, however, and don't have a reference book on hand.

4. Estimation with Whole Numbers: Compatible Numbers

This is changing the numbers in the problem to make them easier to use. For example, if you were adding 348 + 567, you could just take 2 away from the 567, leaving 565, and add it to the 348 so you would have 350. Now you would be adding 350 + 565, and that is a little easier to work with.

For that same problem, you could subtract 52 from the 567, leaving 515, and add it to the 348, to get 400. Then you would have 400 + 517, which is a LOT easier to add.

To subtract, you adjust the amounts in a similar manner... for 567 - 348, you want to avoid regrouping, so add 2 to both amounts... now you have 569 - 350.

The important thing is to make teachers aware that there is more than one way to estimate, and that estimation should be done for a reason, so that the intent is clear, and the problem solving drives the effort.

-Gail, for the T2T service

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